Does $A\geq (\mathrm{tr}(A^{-1}))^{-1}I$ hold for symmetric positive-definite matrix $A$?

Question

Somewhere in my reading, it seems that the following inequality holds for every symmetric positive definite matrix $A$, $$A\geq \big(\mathrm{tr}(A^{-1})\big)^{-1}I,$$ where $I$ is the identity matrix.

Is this true?

Thank you!

Answer 1

Yes. By a change of orthonormal basis, you may assume that $A=\operatorname{diag}(\lambda_1,\ldots,\lambda_n)$ and the inequality becomes $\operatorname{tr}(A^{-1})A\ge I$ or $\lambda_i\sum_{k=1}^n\frac{1}{\lambda_k}\ge1$ for every $i$.

Answer 2

One easy way to see this result is as follows. For any unit vector $v$,

$1 = v^TIv = v^TA^{-1/2} A^{1/2} v \leq \|A^{-1/2} v\|_2 \|A^{1/2} v\|_2 \leq \sqrt{\|A^{-1}\|_{\rm op}} \|A^{1/2}v\|_2 $.

Squaring both sides, $(\|A^{-1}\|_{\rm op})^{-1} v^TIv \leq v^T A v$.

This implies the inequality $$ A \succeq \frac{1}{\|A^{-1}\|_{\rm op}} I. $$ This is stronger than the inequality you were seeking to prove.